javascript - 流程，如何继承类函数类型？

Question

我做了一个简单的路由器类，简化后看起来像这样

// @flow
import { Container } from 'unstated'

type State = {
  history: Object[],
  pathname: string,
  data: Object | null
}

class RouterState extends Container<State> {
  state = {
    history: [{ pathname: '/onboarding/sign-in', data: null }],
    pathname: '/onboarding/sign-in',
    data: null
  }


  mutationPush = (pathname: string, data?: Object | null = null) => {
    if (pathname !== this.state.pathname) {
      const history = [...this.state.history, { pathname, data }]
      this.setState({
        history,
        pathname,
        data
      })
    }
  }
}
export default RouterState

不幸的是 RouterState.mutationPush 不返回类型，因为它不是静态函数，这是预期的，因为 extends Container 的使用以及它在我的 React 中的使用方式应用程序。我正在尝试找出一种方法，可以像这样在类中导出每个单独函数的类型。

编辑具体用例为

type Props = {
  push: RouterState.mutationPush
}

那个错误

[flow] Cannot get RouterState.mutationPush because property mutationPush is missing in statics of RouterState [1]. (References: [1]) [Flow] mutationPush: any

Edit2:我尝试在类中添加类型导出

export type mutationPush: (pathname: string, data?: ?Object) => void

/* ... */

mutationPush: mutationPush = (pathname, data) => ...

但是无论我在哪里使用这个函数，流程助手都会说这个 mutationPush: mutationPush 而不是 mutationPush: (pathname: string, data?: Object) ...

Answer 1

您可以使用 $PropertyType实用程序类型以获取属性或方法的类型。

type Props = {
  push: $PropertyType<RouterState, 'mutationPush'>
}